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frez [133]
3 years ago
13

How much force is needed to accelerate a 7.1 kg skier at 4 m/s/s?

Physics
2 answers:
zysi [14]3 years ago
5 0

Answer: 28.4

Explanation:

F=MA

F=(7.1)(4)

F= 28.4 N

liraira [26]3 years ago
4 0
You would need 28.4 N to accelerate
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A nonconducting sphere of diameter 10.0 cm carries charge distributed uniformly inside with charge density of +5.50 µC/m3 . A pr
VLD [36.1K]

Answer:

t = 2.58*10^-6 s

Explanation:

For a nonconducting sphere you have that the value of the electric field, depends of the region:

rR:\\\\E=k\frac{Q}{r^2}

k: Coulomb's constant = 8.98*10^9 Nm^2/C^2

R: radius of the sphere = 10.0/2 = 5.0cm=0.005m

In this case you can assume that the proton is in the region for r > R. Furthermore you use the secon Newton law in order to find the acceleration of the proton produced by the force:

F=m_pa\\\\qE=m_pa\\\\k\frac{qQ}{r^2}=m_pa\\\\a=k\frac{qQ}{m_pr^2}

Due to the proton is just outside the surface you can use r=R and calculate the acceleration. Also, you take into account the charge density of the sphere in order to compute the total charge:

Q=\rho V=(5.5*10^{-6}C/m^3)(\frac{4}{3}\pi(0.05m)^3)=2.87*10^{-9}C\\\\a=(8.98*10^9Nm^2/C^2)\frac{(1.6*10^{-19}C)(2.87*10^{-9}C)}{(1.67*10^{-27}kg)(0.05m)^2}=9.87*10^{11}\frac{m}{s^2}

with this values of a you can use the following formula:

a=\frac{v-v_o}{t}\\\\t=\frac{v-v_o}{a}=\frac{2550*10^3m/s-0m/s}{9.87*10^{11}m/s^2}=2.58*10^{-6}s

hence, the time that the proton takes to reach a speed of 2550km is 2.58*10^-6 s

3 0
3 years ago
What is law of conservation of momentum ?​
Black_prince [1.1K]

Answer:

conservation of momentum, general law of physics according to which the quantity called momentum that characterizes motion never changes in an isolated collection of objects; that is, the to

8 0
3 years ago
HELPPPP
Arisa [49]

Answer:

A. nuclear fusion reactions

C. it's still hot from the big bang

Explanation:

The inside of the earth is hot due to some reasons. This heat provides the internal energy the drives processes within the earth interior. Here are some of the ways in which the heat has accumulated:

  • Nuclear reactions within the earth interior by fusion and other radioactive processes releases a large amount of heat.
  • Some heat accreted during the early formation of the earth and has not been lost till this day.
  • Heating due to friction

These are some of the sources of the earth's internal heat.

3 0
3 years ago
A 2.5 kg tribble is placed in a bucket and whirled in a 1.4 m radius vertical circle at a constant tangential speed. If the forc
Over [174]

Given that,

Mass of a tribble, m = 2.5 kg

Radius, r = 1.4 m

The force on the tribble from the bucket does not exceed 10 times its weight.

To find,

The maximum tangential speed.

Solution,

The force acting on the tribble is equal to the centripetal force.

F = 10mg

The formula for the centripetal force is given by :

F=\dfrac{mv^2}{r}

v is maximum tangential speed

v=\sqrt{\dfrac{Fr}{m}} \\\\v=\sqrt{\dfrac{mgr}{m}} \\\\v=\sqrt{{10gr}} \\\\v=\sqrt{10\times 9.8\times 1.4} \\\\v=11.7\ m/s

So, the maximum tangential speed is 11.7 m/s.

8 0
3 years ago
The pressure in a traveling sound wave is given by the equation ΔP = (1.78 Pa) sin [ (0.888 m-1)x - (500 s-1)t] Find (a) the pre
frozen [14]

Answer:

a) P_m=1.78\ Pa

b) f=79.5775\ Hz

c) \lambda=7.076\ m

d) v=563.06\ m.s^{-1}

Explanation:

<u>Given equation of pressure variation:</u>

\Delta P= (1.78\ Pa)\ sin\ [(0.888\ m^{-1})x-(500\ s^{-1})t]

We have the standard equation of periodic oscillations:

\Delta P=P_m\ sin\ (kx-\omega.t)

<em>By comparing, we deduce:</em>

(a)

amplitude:

P_m=1.78\ Pa

(b)

angular frequency:

\omega=2\pi.f

2\pi.f=500

∴Frequency of oscillations:

f=\frac{500}{2\pi}

f=79.5775\ Hz

(c)

wavelength is given by:

\lambda=\frac{2\pi}{k}

\lambda=\frac{2\pi}{0.888}

\lambda=7.076\ m

(d)

Speed of the wave is gives by:

v=\frac{\omega}{k}

v=\frac{500}{0.888}

v=563.06\ m.s^{-1}

8 0
3 years ago
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